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%I #12 Jan 22 2017 21:52:00
%S 0,0,1,2,0,1,1,2,2,0,3,4,1,1,1,2,2,2,6,4,0,3,3,4,4,1,5,5,1,1,1,2,2,2,
%T 7,7,2,6,6,4,4,0,5,8,3,3,3,4,4,4,9,6,1,5,5,5,5,1,6,6,1,1,1,2,2,2,8,8,
%U 2,7,7,7,7,2,8,12,6,6,6,4,4,4,10,6,0,5,5,8,8,3,9,9,3,3,3
%N Write n in binary reflected Gray code and sum the positions where there is a '1' followed immediately to the right by a '0', counting the leftmost digit as position 1.
%H Indranil Ghosh, <a href="/A281388/b281388.txt">Table of n, a(n) for n = 1..10000</a>
%F a(n) = A049501(A003188(n)).
%e For n = 11, the binary reflected Gray code for 11 is '1110'. In '1110', the position of '1' followed immediately to the right by '0' counting from left is 3. So, a(11) = 3.
%e For n = 12, the binary reflected Gray code for 12 is '1010'. In '1010', the positions of '1' followed immediately to the right by '0' counting from left are 1 and 3. So, a(12) = 1 + 3 = 4.
%o (Python)
%o def g(n):
%o ....return bin(n^(n/2))[2:]
%o def a(n):
%o ....x=g(n)
%o ....s=0
%o ....for i in range(1,len(x)):
%o ........if x[i-1]=="1" and x[i]=="0":
%o ............s+=i
%o ....return s
%Y Cf. A003188, A014550, A049501.
%K nonn,base
%O 1,4
%A _Indranil Ghosh_, Jan 21 2017
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